I agree that you can make a betting/scoring setup such that betting/predicting at 50% is correct. Eg,suppose that on both Monday and Tuesday, Beauty gets to make a $1 bet at some odds. If she’s asleep on Tuesday then whatever bet she made on Monday is repeated. In that case she should bet at 1:1 odds.
Let’s work it out if Beauty has 1⁄3 probability for Heads, with the 2⁄3 probability for Tails split evenly between 1⁄3 for Tails&Monday and 1⁄3 for Tails&Tuesday.
Here, the possible actions are “don’t bet” or “bet on Heads”, at 1:1 odds (let’s say win $1 or lose $1), on each day for which a bet is active.
The expected gain from not betting is of course zero. The expected gain from betting on some day can be computed by summing the gains in each situation (Heads&Monday, Tails&Monday, Tails&Tuesday) times the corresponding probabilities:
P(Heads&Monday) x 2 + P(Tails&Monday) x (-1) + P(Tails&Tuesday) x (-1)
This evaluates to zero. Note that the gain for betting in the Heads&Monday situation is $2 because if she makes this bet, it is active for two days, on each of which it wins $1. Since the expected gain is zero for both betting and not betting, Beauty is indifferent to betting on Heads or not betting. If the odds shifted a bit, so winning the bet paid $1 but losing the bet cost $1.06, then the expected gain would be minus $0.04, and Beauty should not make the bet on Heads.
Note that when Beauty is woken on both Monday and Tuesday, we expect that she will make the same decision both days, since she has no reason to make different decisions, but this is not some sort of logical guarantee—Beauty is a human being, who makes decisions for herself at each moment in time, which can conceivably (but perhaps with very low probability) be inconsistent with other decisions she makes at other times.
Now suppose that Beauty’s probability for Heads is 1⁄2, so that P(Heads&Monday)=1/2, P(Tails&Monday)=1/4, and P(Tails&Tuesday)=1/4, though the split between Tails&Monday and Tails&Tuesday doesn’t actually matter in this scenario.
Plugging these probabilities into the formula above, we compute the expected gain to be $0.50. So Beauty is not indifferent, but instead will definitely want to bet on Heads. If we shift the payoffs to win $1 lose $1.06, then the expected gain is $0.47, so Beauty will still want to bet on Heads.
However, when a win pays $1 and a loss costs $1.06, arguments from an outside perspective say that if every day she is woken Beauty follows this strategy of betting on Heads, her expected gain is minus $0.03 per experiment. [Edit: Correction, the expected gain per experiment is minus $0.06 - I forgot to multiply by two for the bet being made twice.]
So we see that Beauty makes a mistake if her probability for Heads is 1⁄2. She makes the right decision (if she applies standard decision theory) if her probability of Heads is 1⁄3.
Note that this is about how much money Beauty gets. One can get the right answer by applying standard decision theory. There is no role for some arbitrary “scoring setup”.
I agree that you can make a betting/scoring setup such that betting/predicting at 50% is correct. Eg,suppose that on both Monday and Tuesday, Beauty gets to make a $1 bet at some odds. If she’s asleep on Tuesday then whatever bet she made on Monday is repeated. In that case she should bet at 1:1 odds.
Let’s work it out if Beauty has 1⁄3 probability for Heads, with the 2⁄3 probability for Tails split evenly between 1⁄3 for Tails&Monday and 1⁄3 for Tails&Tuesday.
Here, the possible actions are “don’t bet” or “bet on Heads”, at 1:1 odds (let’s say win $1 or lose $1), on each day for which a bet is active.
The expected gain from not betting is of course zero. The expected gain from betting on some day can be computed by summing the gains in each situation (Heads&Monday, Tails&Monday, Tails&Tuesday) times the corresponding probabilities:
P(Heads&Monday) x 2 + P(Tails&Monday) x (-1) + P(Tails&Tuesday) x (-1)
This evaluates to zero. Note that the gain for betting in the Heads&Monday situation is $2 because if she makes this bet, it is active for two days, on each of which it wins $1. Since the expected gain is zero for both betting and not betting, Beauty is indifferent to betting on Heads or not betting. If the odds shifted a bit, so winning the bet paid $1 but losing the bet cost $1.06, then the expected gain would be minus $0.04, and Beauty should not make the bet on Heads.
Note that when Beauty is woken on both Monday and Tuesday, we expect that she will make the same decision both days, since she has no reason to make different decisions, but this is not some sort of logical guarantee—Beauty is a human being, who makes decisions for herself at each moment in time, which can conceivably (but perhaps with very low probability) be inconsistent with other decisions she makes at other times.
Now suppose that Beauty’s probability for Heads is 1⁄2, so that P(Heads&Monday)=1/2, P(Tails&Monday)=1/4, and P(Tails&Tuesday)=1/4, though the split between Tails&Monday and Tails&Tuesday doesn’t actually matter in this scenario.
Plugging these probabilities into the formula above, we compute the expected gain to be $0.50. So Beauty is not indifferent, but instead will definitely want to bet on Heads. If we shift the payoffs to win $1 lose $1.06, then the expected gain is $0.47, so Beauty will still want to bet on Heads.
However, when a win pays $1 and a loss costs $1.06, arguments from an outside perspective say that if every day she is woken Beauty follows this strategy of betting on Heads, her expected gain is minus $0.03 per experiment. [Edit: Correction, the expected gain per experiment is minus $0.06 - I forgot to multiply by two for the bet being made twice.]
So we see that Beauty makes a mistake if her probability for Heads is 1⁄2. She makes the right decision (if she applies standard decision theory) if her probability of Heads is 1⁄3.
Note that this is about how much money Beauty gets. One can get the right answer by applying standard decision theory. There is no role for some arbitrary “scoring setup”.